\newproblem{lay:4_1_32}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.1.32}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $H$ and $K$ be subspaces over a vector space $V$. The intersection of $H$ and $K$, written as $H\cap K$, is the set of all vectors $\mathbf{v}\in V$
	that belong to both $H$ and $K$. Show that $H\cap K$ is a subspace of $V$. (See figure below.) Give an example in $\mathbb{R}^2$ to show that the union of
	subspaces is not, in general, a subspace.
}{
  % Solution
	We need to show that this $H\cap K$ meets the three requirements to be a subspace
	\begin{itemize}
		\item $\mathbf{0}\in H\cap K$ \\
		      This is true because for $\mathbf{0}$ belongs to both $H$ and $K$ since both of them are, in their turn, subspaces.
		\item Given any two vectors $\mathbf{u},\mathbf{v}\in H\cap K$, $\mathbf{u}+\mathbf{v}\in H\cap K$ \\
		      $\mathbf{u}$ and $\mathbf{v}$ belong to both $H$ and $K$. And these sets are subspaces, then
					\begin{center}
						$\mathbf{u}+\mathbf{v}\in H$ \\
						$\mathbf{u}+\mathbf{v}\in K$ \\
					\end{center}
					So $\mathbf{u}+\mathbf{v}\in H\cap K$
		\item Given any vector $\mathbf{u}\in H\cap K$ and $c\in\mathbb{R}$, $c\mathbf{u}\in H\cap K$ \\
		      $\mathbf{u}$ belongs to both $H$ and $K$. And these sets are subspaces, then
					\begin{center}
						$c\mathbf{u}\in H$ \\
						$c\mathbf{u}\in K$ \\
					\end{center}
					So $c\mathbf{u}\in H\cap K$
	\end{itemize}
	Since $H\cap K$ meets all properties, $H\cap K$ is a subspace of $V$.\\
	The union of subspaces is not, in general, a subspace. For instance in $\mathbb{R}^2$, the following sets are subspaces:
	\begin{center}
		$H=\{(x,0)\in\mathbb{R}^2\}$ \\
		$K=\{(0,y)\in\mathbb{R}^2\}$ \\
	\end{center}
	but the union $H\cup K$ is not a subspace. For instance, $\mathbf{u}=(1,0)\in H\cup K$ and $\mathbf{v}=(0,1)\in H\cup K$, but
	$\mathbf{u}+\mathbf{v}=(1,1)\notin H\cup K$.
}
\useproblem{lay:4_1_32}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
